# Body of mass m is accelerating -- Need to find resultant force

1. Oct 31, 2016

### dokara97

1. Body of mass m = 3kg is accelerating with acceleration of $\vec a = 2\vec i + 5\vec j m/s^2$. I need to find the resultant force $\vec F$ and the force's value.

2. Of course, the first equation that came to my mind is $\vec F = m * \vec a$

3. I came with a possible solution for this task pretty fast, but it just seems too easy. I was thinking to find the norm of this vector and just calculate the force by the equation I just wrote up there. But as I said it just seems way to easy. I would put that vector in at graph and just assume it starts from T(0,0) and I could easily calculate the norm of that vector. I don't have the solution so, I am blindfolded. This is my first thread with problem in it, so I hope I respected all the rules

Last edited: Oct 31, 2016
2. Oct 31, 2016

### Staff: Mentor

Hi dokara97, Welcome to Physics Forums.

The problem really is as simple as you thought, and your relevant equation applies. By the "force's value" I assume that you mean the magnitude of the force, so yes, the norm is the way to go (To be mathematically precise, it's what's known as the L2-norm).

3. Oct 31, 2016

### dokara97

By that you mean that text that was written by 1.2.3. numbers? If it is that, than it's at my knowledge. Hmm, my task asks for resultant force $\vec F$ and it's ''amount'' . I am translating it literally from my language as it says in the task. So with L2-norm I can find the amount of that force I suppose. But what's the deal with the resultant force? Isn't it the same thing? I dont have component's of the force to calculate the resultant force by adding components together. Sorry if I am not writing some things 100% properly, I am really trying my best to describe in English the problems from my language.

4. Oct 31, 2016

### Staff: Mentor

Yes, the template headers are provided automatically in the edit window. They are numbered and in bold font. They should be kept in place and used to format your question.

You are given the acceleration vector and the mass. The force vector is $\vec F = m * \vec a$ as you wrote. That will be the resultant force. The acceleration of an object is due to the resultant of all forces acting on it, so automatically if you use the acceleration to find the force, you've found the resultant force.

I believe that they want you to give the force in component form and determine the magnitude of that force.

5. Oct 31, 2016

### dokara97

In future I wont remove template headers. Can you please define the ''magnitude'' how is it different from resultant force. When I translate magnitude, I get the same translation as size. How is it different from what I get by calculating the force with that equation? Im sorry if my questions are dumb or something, but I am having some problems with terminology currently, but activity on this community will help me alot with that.

6. Oct 31, 2016

### Staff: Mentor

The force is a vector. It has both magnitude and direction. The magnitude is its size. We typically calculate it using the L2-norm equation: $|\vec F| = \sqrt{F_x^2 + F_y^2 + F_z^2}$

The magnitude of the force is a scalar. It has the units of force (N for example) but no direction.

When you read something like: "A 30 N force is applied at an angle of 35 degrees to the horizontal...", then 30 N is its magnitude and 35 degrees is directional information.

7. Nov 1, 2016

### dokara97

Thank you, now you made it much more clear to understand for me.

8. Nov 1, 2016

### dokara97

So, just to make myself sure before I go and explain my homework in class and get graded, direction of $\vec F$ will be pointed from XY(0,0) to $\vec a = 2\vec i + 5\vec j m/s^2$ ? I mean, that is the only way that seems logic because, the $\vec a$ depends on $\vec F$, but not the opposite.

9. Nov 1, 2016

### Staff: Mentor

I think you have the right notion but are being a bit casual with the details. XY(0,0) would be a location in space whereas $\vec a$ is a vector that can be placed anywhere. Wherever it is placed it will have the same magnitude and direction.

So a better description of what you want to convey is that the force vector will have the same direction as the acceleration vector and that direction is the same as that of a position vector from the origin XY(0,0) to point XY(2,5). Or, if you will, it will have the same direction as a vector $\vec r = 2\vec i + 5\vec j m$.

Does that help?

10. Nov 1, 2016

### dokara97

Yes, I forgot to say that what I meant by force vector is radius vector from XY(0,0) to XY(2,5). You definitely made me aware to be careful with the terms and be more specific so, thanks again.

11. Nov 2, 2016

### dokara97

Im sorry, but I noticed what you wrote right now. So when I calculate all of this in the way we discussed here, what else should I do to give the force in component form and determine the magnitude of that force. The magnitude as you said is the same as the magnitude of acceleration, but how can I give the force in component form? So you said the $\vec F$ will have the same direction as vector $\vec r = 2\vec i + 5\vec j$ so can I just give that vector in components with parallelogram method and is that going to be the correct solution or?

12. Nov 2, 2016

### Staff: Mentor

First find $\vec F = m \vec a$. That will give you the force vector in component form. I assume that you know how to multiply a vector by a scalar? Then find $| \vec F |$, its magnitude.

The magnitude of the force vector is not the same as the magnitude of the acceleration vector; the force vector is the mass times the acceleration vector, so their magnitudes will be scaled accordingly. Also, they have different units. It's the directions of the two vectors that are the same.

13. Nov 6, 2016

### dokara97

Thank you very much. Sorry for late reply but I had been busy alot last days. The problem I got is indeed very easy, but I needed to solve it on various ways, by that I mean to prove all of my results in all sort of ways, mostly vectors.